# Math Help - Derive Logistic Growth Equation

1. ## Derive Logistic Growth Equation

Hey guys. I'm really stuck on this homework. The instructions were "Dervie the logistic growth equation".

So, the teacher gave us the following stuff:

dy/dt = ky (M-y)
k, t > 0

y(0) = M / 1+c

y = M / 1+ce ^ -kMt

So, i'm guessing I need to dervie y = M / 1+ce ^ -kMt right?

This is how i went about it.

y = M / 1+ce ^ -kMt
dy/dt = (1+ce ^ -kMt)(M derivative) - (M)(-kMt ce ^ -kMt-1)(ce derivative) / (1+ce ^ -kMt)^2

2. Hello, anfieldgirl!

This is a Differential Equation problem.
. . We must solve it accordingly.

Derive the logistic growth equation.

. . $\frac{dy}{dt} \:=\: ky (M-y),\;\;k, t > 0,\;\;y(0) \:=\:\frac{M}{1+c}$

Answer: . $y \:= \:\frac{M}{1+ce^{-kMt}}$

We have: . $\frac{dy}{dt} \:=\:ky(M-y)$

Separate variables: . $\frac{dy}{y(M-y)} \:=\:k\,dt$

On the left, use Partial Fraction Decomposition:

. . $\frac{1}{M}\left(\frac{1}{y} + \frac{1}{M-y}\right)dy \:=\:k\,dt$

Integrate: . $\frac{1}{M}\left[\ln y - \ln(M-y)\right] \;=\;kt + C_1 \quad\Rightarrow\quad \ln\left(\frac{y}{M-y}\right) \:=\:kMt + C_2$

. . Then: . $\frac{y}{M-y} \:=\:e^{kMt + C_2} \:=\:e^{kMt}\cdot e^{C_2} \;=\;Ce^{kMt}$

We have: . $y \:=\:Ce^{kMt}(M - y) \quad\Rightarrow\quad y \;=\;CMe^{kMt} - Cye^{kMt}$

. . $Cye^{kMt}+ y \;=\;CMe^{kMt} \quad\Rightarrow\quad y\left(Ce^{kMt} + 1\right) \;=\;CMe^{kMt}
$

Then: . $y \;=\;\frac{CMe^{kMt}}{C^{kMt}+1}$

Divide top and bottom by $e^{kMt}\!:\;\;y \;=\;\frac{CM}{e^{-kMt} + 1}$

Then use the initial condition to evaluate $C.$

3. Thanks so much for replying. I have some questions, i need to clarify though.

First, why do we seperate the variables?

Second, is there a different way other than Partial Fraction Decomposition? Because my teacher didn't teach us that method yet, so i'm confused...

Thank you so much again.